If A,Bε(0,(π/2)) such that 3sin^2 A+2sin^2 B=1 and 3sin2A−2sin2B=0, find the value of A+B.


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Question Number 27820 by bmind4860 last updated on 15/Jan/18

$I f A, B ϵ (0, \frac{π}{2}) s u c h t h a t 3 {s i n}^{2} A + 2 {s i n}^{2} B = 1$ $a n d 3 s i n 2 A - 2 s i n 2 B = 0, f i n d t h e v a l u e o f A + B .$
	Answered by ajfour last updated on 15/Jan/18

	$\frac{3}{2} (1 - \cos 2 A) + 1 - \cos 2 B = 1$ $\Rightarrow 3 \cos 2 A + 2 \cos 2 B = 3 . . . . (i)$ $a n d s i n c e 3 \sin 2 A = 2 \sin 2 B$ $9 \sin^{2} 2 A = 4 \sin^{2} 2 B$ $\Rightarrow 9 - 9 \cos^{2} 2 A = 4 - 4 \cos^{2} 2 B$ $\Rightarrow 9 \cos^{2} 2 A - 4 \cos^{2} 2 B = 5$ $o r (3 \cos 2 A - 2 \cos 2 B) (3 \cos 2 A + 2 \cos 2 B) = 5$ $u s i n g (i) :$ $3 \cos 2 A - 2 \cos 2 B = \frac{5}{3}$ $a n d a s 3 \cos 2 A + 2 \cos 2 B = 3$ $a d d i n g t h e a b o v e t w o e q s .$ $6 \cos 2 A = \frac{14}{3} \Rightarrow \cos 2 A = \frac{7}{9}$ $a n d \sin 2 A = \sqrt{1 - \frac{49}{81}} = \frac{4 \sqrt{2}}{9}$ $s u b t r a c t i n g t h e m,$ $4 \cos 2 B = \frac{4}{3} \Rightarrow \cos 2 B = \frac{1}{3}$ $a n d \sin 2 B = \sqrt{1 - \frac{1}{9}} = \frac{2 \sqrt{2}}{3}$ $\cos (2 A + 2 B) = \cos 2 A \cos 2 B$ $- \sin 2 A \sin 2 B$ $= \frac{7}{9} \times \frac{1}{3} - \frac{4 \sqrt{2}}{9} \times \frac{2 \sqrt{2}}{3}$ $= - \frac{1}{3}$ $\cos (A + B) = \sqrt{\frac{1 + \cos (2 A + 2 B)}{2}}$ $= \frac{1}{\sqrt{3}}$ $A + B = \cos^{- 1} (\frac{1}{\sqrt{3}}) .$
	Commented by bmind4860 last updated on 15/Jan/18

	$p e r f e c t$
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